Integrand size = 98, antiderivative size = 25 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {1}{5 \left (3+x+\frac {e^x x \log \left (x^2\right )}{x+x^2}\right )} \] Output:
1/(5*x+5*exp(x)*x/(x^2+x)*ln(x^2)+15)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=-\frac {-1-x}{5 \left (3+4 x+x^2+e^x \log \left (x^2\right )\right )} \] Input:
Integrate[(E^x*(-2 - 2*x) - x - 2*x^2 - x^3 - E^x*x^2*Log[x^2])/(45*x + 12 0*x^2 + 110*x^3 + 40*x^4 + 5*x^5 + E^x*(30*x + 40*x^2 + 10*x^3)*Log[x^2] + 5*E^(2*x)*x*Log[x^2]^2),x]
Output:
-1/5*(-1 - x)/(3 + 4*x + x^2 + E^x*Log[x^2])
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-x^3-2 x^2-e^x x^2 \log \left (x^2\right )-x+e^x (-2 x-2)}{5 x^5+40 x^4+110 x^3+120 x^2+5 e^{2 x} x \log ^2\left (x^2\right )+e^x \left (10 x^3+40 x^2+30 x\right ) \log \left (x^2\right )+45 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-(x+1) \left (x^2+x+2 e^x\right )-e^x x^2 \log \left (x^2\right )}{5 x \left (x^2+e^x \log \left (x^2\right )+4 x+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {e^x \log \left (x^2\right ) x^2+(x+1) \left (x^2+x+2 e^x\right )}{x \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {e^x \log \left (x^2\right ) x^2+(x+1) \left (x^2+x+2 e^x\right )}{x \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {\log \left (x^2\right ) x^2+2 x+2}{x \log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )}-\frac {(x+1) \left (\log \left (x^2\right ) x^3+2 \log \left (x^2\right ) x^2+2 x^2-\log \left (x^2\right ) x+8 x+6\right )}{x \log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-\int \frac {1}{\left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+\int \frac {x}{\left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+3 \int \frac {x^2}{\left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+14 \int \frac {1}{\log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+6 \int \frac {1}{x \log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+10 \int \frac {x}{\log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx+2 \int \frac {x^2}{\log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx-\int \frac {x}{x^2+4 x+e^x \log \left (x^2\right )+3}dx-2 \int \frac {1}{\log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )}dx-2 \int \frac {1}{x \log \left (x^2\right ) \left (x^2+4 x+e^x \log \left (x^2\right )+3\right )}dx+\int \frac {x^3}{\left (x^2+4 x+e^x \log \left (x^2\right )+3\right )^2}dx\right )\) |
Input:
Int[(E^x*(-2 - 2*x) - x - 2*x^2 - x^3 - E^x*x^2*Log[x^2])/(45*x + 120*x^2 + 110*x^3 + 40*x^4 + 5*x^5 + E^x*(30*x + 40*x^2 + 10*x^3)*Log[x^2] + 5*E^( 2*x)*x*Log[x^2]^2),x]
Output:
$Aborted
Time = 0.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {2+2 x}{10 \,{\mathrm e}^{x} \ln \left (x^{2}\right )+10 x^{2}+40 x +30}\) | \(25\) |
risch | \(\frac {\frac {2}{5}+\frac {2 x}{5}}{-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}+2 x^{2}+4 \,{\mathrm e}^{x} \ln \left (x \right )+8 x +6}\) | \(79\) |
Input:
int((-x^2*exp(x)*ln(x^2)+(-2-2*x)*exp(x)-x^3-2*x^2-x)/(5*x*exp(x)^2*ln(x^2 )^2+(10*x^3+40*x^2+30*x)*exp(x)*ln(x^2)+5*x^5+40*x^4+110*x^3+120*x^2+45*x) ,x,method=_RETURNVERBOSE)
Output:
1/10*(2+2*x)/(exp(x)*ln(x^2)+x^2+4*x+3)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {x + 1}{5 \, {\left (x^{2} + e^{x} \log \left (x^{2}\right ) + 4 \, x + 3\right )}} \] Input:
integrate((-x^2*exp(x)*log(x^2)+(-2-2*x)*exp(x)-x^3-2*x^2-x)/(5*x*exp(x)^2 *log(x^2)^2+(10*x^3+40*x^2+30*x)*exp(x)*log(x^2)+5*x^5+40*x^4+110*x^3+120* x^2+45*x),x, algorithm="fricas")
Output:
1/5*(x + 1)/(x^2 + e^x*log(x^2) + 4*x + 3)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {x + 1}{5 x^{2} + 20 x + 5 e^{x} \log {\left (x^{2} \right )} + 15} \] Input:
integrate((-x**2*exp(x)*ln(x**2)+(-2-2*x)*exp(x)-x**3-2*x**2-x)/(5*x*exp(x )**2*ln(x**2)**2+(10*x**3+40*x**2+30*x)*exp(x)*ln(x**2)+5*x**5+40*x**4+110 *x**3+120*x**2+45*x),x)
Output:
(x + 1)/(5*x**2 + 20*x + 5*exp(x)*log(x**2) + 15)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {x + 1}{5 \, {\left (x^{2} + 2 \, e^{x} \log \left (x\right ) + 4 \, x + 3\right )}} \] Input:
integrate((-x^2*exp(x)*log(x^2)+(-2-2*x)*exp(x)-x^3-2*x^2-x)/(5*x*exp(x)^2 *log(x^2)^2+(10*x^3+40*x^2+30*x)*exp(x)*log(x^2)+5*x^5+40*x^4+110*x^3+120* x^2+45*x),x, algorithm="maxima")
Output:
1/5*(x + 1)/(x^2 + 2*e^x*log(x) + 4*x + 3)
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {x + 1}{5 \, {\left (x^{2} + e^{x} \log \left (x^{2}\right ) + 4 \, x + 3\right )}} \] Input:
integrate((-x^2*exp(x)*log(x^2)+(-2-2*x)*exp(x)-x^3-2*x^2-x)/(5*x*exp(x)^2 *log(x^2)^2+(10*x^3+40*x^2+30*x)*exp(x)*log(x^2)+5*x^5+40*x^4+110*x^3+120* x^2+45*x),x, algorithm="giac")
Output:
1/5*(x + 1)/(x^2 + e^x*log(x^2) + 4*x + 3)
Time = 2.89 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {\frac {2\,x\,{\mathrm {e}}^{2\,x}}{5}+\frac {2\,x^2\,{\mathrm {e}}^{2\,x}}{5}-{\mathrm {e}}^x\,\left (\frac {x^5}{5}+\frac {3\,x^4}{5}+\frac {x^3}{5}-\frac {x^2}{5}\right )}{\left (4\,x+\ln \left (x^2\right )\,{\mathrm {e}}^x+x^2+3\right )\,\left (2\,x\,{\mathrm {e}}^{2\,x}+x^2\,{\mathrm {e}}^x-2\,x^3\,{\mathrm {e}}^x-x^4\,{\mathrm {e}}^x\right )} \] Input:
int(-(x + exp(x)*(2*x + 2) + 2*x^2 + x^3 + x^2*log(x^2)*exp(x))/(45*x + 12 0*x^2 + 110*x^3 + 40*x^4 + 5*x^5 + log(x^2)*exp(x)*(30*x + 40*x^2 + 10*x^3 ) + 5*x*log(x^2)^2*exp(2*x)),x)
Output:
((2*x*exp(2*x))/5 + (2*x^2*exp(2*x))/5 - exp(x)*(x^3/5 - x^2/5 + (3*x^4)/5 + x^5/5))/((4*x + log(x^2)*exp(x) + x^2 + 3)*(2*x*exp(2*x) + x^2*exp(x) - 2*x^3*exp(x) - x^4*exp(x)))
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {e^x (-2-2 x)-x-2 x^2-x^3-e^x x^2 \log \left (x^2\right )}{45 x+120 x^2+110 x^3+40 x^4+5 x^5+e^x \left (30 x+40 x^2+10 x^3\right ) \log \left (x^2\right )+5 e^{2 x} x \log ^2\left (x^2\right )} \, dx=\frac {-e^{x} \mathrm {log}\left (x^{2}\right )-x^{2}+1}{20 e^{x} \mathrm {log}\left (x^{2}\right )+20 x^{2}+80 x +60} \] Input:
int((-x^2*exp(x)*log(x^2)+(-2-2*x)*exp(x)-x^3-2*x^2-x)/(5*x*exp(x)^2*log(x ^2)^2+(10*x^3+40*x^2+30*x)*exp(x)*log(x^2)+5*x^5+40*x^4+110*x^3+120*x^2+45 *x),x)
Output:
( - e**x*log(x**2) - x**2 + 1)/(20*(e**x*log(x**2) + x**2 + 4*x + 3))